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For a gaseous mixtue of 2.41g of helium ...

For a gaseous mixtue of `2.41g` of helium and `2.79g` of neon in an evacuated `1.04 dm^(3)` container at 298 K Calculate the partial pressure of each gas and hence find the total pressure of the mixture.

Text Solution

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The correct Answer is:
Mass of He = 2 . 41 g
No. of moles of He =
`("Mass")/("Molar Mass") = (2.41)/(4) = 0 . 6 0 25 ` moles
Mass Of Ne = 2 . 79 g
No. of moles of Ne ` = ("Mass")/("Molar Mass") = (2.79)/(20)`
= 0 . 1395 moles
Volume of the Total no. of moles of the mixture Conysinrt V = 1 . 0 4 `dm^(3)`
Temperture T = 298 K
Pressure P` = (1)/(V) RT`
According to ideal gas equation PV = nRT
` P = ( 0 7 4 2 0 xx 0 . 0 821 xx 298 )/( 1 . 0 4) = 17 . 45` atm
Partial pressure P = molefraction ` xx` Total pressure
` = (nA)/(nA + nB) xx P `
Partial Pressure of Helium ` = P_(He) = (0 . 6025)/(0.7420) xx 17 . 45 `
According to Dalton's law of partial pressure = 3 . 280 atm
`P = P_(1) + P_(2) + P_(3). . . . `
` P_("total") = P_(He) + P_(Ne) = 14. 169 + 3 . 280 = 17 . 449 ` atm
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